On the problems of edge disjoint cliques in graphsThesis

Nguyen Huu Hoi

Supervisor: Gyori Ervin

2005

Eotvos Lorand University, Mathematics Department

Table of contents

Preface. 2

Chapter 1. Estimates on general graphs. 4

Chapter 2. Estimates on graphs of small size. 15

Chapter 3. Other methods and relating problems. 24

References. 33

1

Preface

Many results have been published in extremal graph theory on the number of given sizecliques in graphs. To some extent, there are three main topics:

• on the maximal number of cliques.

• on the maximal number of vertex disjoint cliques.

• on the maximal number of edge disjoint ones.

For the first topic, Moon and Moser [11] proved that a graph of order n and size e (i.e ithas n vertices and e edges) contains at least (e/3n)(4e−n2) triangles, the analogous resultscan be proved for the general case. If e is bounded from above then better lower boundscan be proved, say for the number of triangles, Bollobas [2] proved that if n2/4 ≤ e ≤ n2/3then the graph contains at least (n/9)(4e− n2) triangles.

From another viewpoint, Erdos conjectured that if G is a graph of order n and sizebn2/4c+m (m < n/2) then G has at least mbn/2c triangles. Lovasz and Simonovits [12]proved this conjecture whenever n is sufficiently large. More results can be found in [3],[4] and [14].

Dealing with the second one, Hajnal and Szemeredi [1] showed that if the minimumdegree δ(G) ≥ (1 − 1/r)n then G contains bn/rc vertex-disjoint copies of Kr. One canreplace Kr by a H graph of order r, and the extended version of the Hajnal-Szemeredibecomes a theorem of Alon and Yuster [15] concerning the vertex disjoint copies of Hwhen δ(G) is greater than (1− 1/χ(H))n. These problems are studied in a large numberof papers by E. Szemeredi, J. Komlos, M. Simonovits, R. Yuster.

The thesis concentrates on the third topic, studying on the number of edge disjoint cliques.For a graph G of order n and size m (which are given), let edkk(G) denote the maximumnumber of edge disjoint Kk’s in G. Let edkk(n,m) denote the minimum of edkk(G) forall graphs G of order n and size m.

2

In the first chapter, generally speaking, we will prove edk3(n, n2

4+ m) ≥ 5

9m − O(n),

edk4(G) ≥ 518

m− o(n2) and edkk(n, tk−1,n + m) ≥ 1

(k2)−(k−2)

m for k ≥ 5. The latter could

be refined by method applied in the first two, however the result would not be muchbetter and still far from the possibly best edkk(G) ≥ 2

km− o(n2).

The second chapter discusses the case the graph contains tr−1(n) + m edges, wherem = o(n2). There are almost m cliques can be found. Generally,

edkk(n, tk−1(n) + m) = m−O(m2

n2)

andedkk(n, tk−1(n) + m) = m

if m is ”linear”.

The last chapter is miscellaneous and contains three small topics. The first two deala bit with fractional method. On the third we are likely to have some notes upon therelation between edkk(G) and other structures of the graph.

ThanksI am very grateful to my advisor Gyori Ervin for giving me this pretty topic. Withouthis help and encouragement I could not have finished the work.

3

Chapter 1

For n = t(k − 1) + r, 0 ≤ r ≤ k − 2, let Tk−1,n denote the complete (k-1)-partite graphof n vertices such that r color classes contain t + 1 vertices and k − r − 1 color classescontain t vertices. Then Tk−1,n, which has tk−1,n := |E(Tk−1,n)| = [n(n− t)− r(t + 1)]/2edges, does not contain any complete subgraph Kk of k vertices. Turan proved that anyother graph of n vertices with at least tk−1,n edges contains Kk as a subgraph. Tk−1,n iscalled the (k − 1)−partite Turan graph.Results of Erdos, Goodman and Posa [16] (k = 3) and Bollobas [2] (k ≥ 4) imply that

T 1. The edge set of every graph of n vertices can be decomposed into at most tk−1,n edgedisjoint Kk’s and edges. If k ≥ 3 then Tk−1,n is the only extremal graph. For k = 3, thecomplete graph K4, K5 and the graphs T2,n(n = 1, 2, ..) are the extremal graphs.

The results can be formulated in the following form

edkk(n, tk−1,n + m) ≥ 1

(k2)−1

m if k ≥ 3.

In a paper by Gyori and Tuza [6], the result was improved (k ≥ 4).Let pk(G) =min{∑m

1 |V (Gi)| : Gi’s are Kk and edges,∪mi E(Gi) = E(G)}. If G is Kk−free

then pk(G) = 2|E(G)| and, in particular, pk(Tk−1,n) = 2tk−1,n. The authors stated that

T 2. If k ≥ 4 and G is an arbitrary graph of n vertices then pk(G) ≤ 2tk−1,n and equalityholds only for Tk−1,n.

Which means

edkk(n, tk−1,n + m) ≥ 1

(k2)− k

2

m if k ≥ 4

equality holds only for Turan graphs.(k ≥ 4 is supposed because p3(G) is not always smaller than 2t2(n). If we cover thecomplete graph K6m−2 by edge disjoint triangles and edges, the number of edges used inthe cover is at least 3m, so, p3(K6m−2) ≥ |E(K6m−2)|+ 3m = 2t2,6m−2 + 1)By using the same method, the result can be improved a bit.

4

Theorem 1.1. (n ≥ k ≥ 3)

edkk(n, tk−1,n + m) ≥ 1(k2

)− (k − 2)m

and only Tk−1,n, for k = 3, K4,K5, T2,n are the extremal examples.

To prove this, we need the a result of Hajnal and Szemeredi[1] which has been noticedin the preface.

T 3. Let G be a graph of n vertices such that the degree of every vertex in G is atleast d. Then there exist vertex disjoint complete subgraphs G1,...,Gn−d of G such that∪n−d

1 V (Gi) = V (G) and the numbers of the vertices of the subgraphs Gi differ from eachother at most one, i.e,|V (Gi)| is b n

n−dc or d n

n−de for 1 ≤ i ≤ n− d.

Proof of Theorem 1.1. The case k = 3 was settled. We prove for k ≥ 4 by induction onn.If n = k the statement holds obviously. Suppose that it holds for graphs of at most n− 1vertices.Let

n− 1 = t(k − 1) + r, 0 ≤ r ≤ k − 2, t ≥ 1.

Thentk−1,n − tk−1,n−1 = n− t− 1 = t(k − 2) + r.

Consider an arbitrary graph G of n vertices and tk−1(n) + m edges. Let x be a vertex ofG with minimum degree, d := dG(x).If d ≤ n− 1− t, then

edkk(G) ≥ edkk(G− x) ≥ 1(k2

)− (k − 2)(tk−1,n + m− d− tk−1,n−1)

≥ 1(k2

)− (k − 2)m.

Equality holds iff d = n− t− 1, G− x = Tk−1,n−1, and m = 0, so G = Tk−1,n (otherwiseTuran’s theorem implies edkk(G) ≥ 1)From now on, suppose

d ≥ n− t = t(k − 2) + r + 1.

In this case we will prove the strict inequality.Because of the minimality of d(x), the degree of every vertex in G(N(x))(the inducedgraph by neighbors of x) is at least 2d− n.Applying T 3 to G(N(x)), we obtain that there exist vertex disjoint complete subgraphs

5

G1,..,Gn−d of order b dn−d

c or d dn−d

e and ∪n−d1 V (Gi) = V (N(x)).

Sinced

n− d≥ t(k − 2) + r + 1

t> k − 2

There are two cases to be considered:

Case 1. There exists Gi with |V (Gi)| = k − 2.Suppose |V (Gi)| = k − 1 for i = 1, ..., p and |V (Gi)| = k − 2 for i = p + 1, ..., n− d wherep ≤ n− d− 1.

d = p(k − 1) + (n− p− d)(k − 2)

implies thatp = d(k − 1)− n(k − 2). (1.1)

Note that p ≥ 1 (otherwise, d = n(k−2)k−1

and thus dn−d

= k − 2, contradiction.)

The graph H := (V (G) − x,E(G)\ ∪p1 E(Gi)) has tk−1,n + m − d − p

(k−12

)edges. By

the induction hypothesis,

edkk(G) ≥ p + edkk(H) ≥ p +1(

k2

)− (k − 2)(tk−1,n + m− d− p

(k − 1

2

)− tk−1,n−1)

=1(

k2

)− (k − 2)m +

1(k2

)− (k − 2)(p + t(k − 2) + r − d) >

1(k2

)− (k − 2)m

Becausep + t(k − 2) + r − d = p− 1 + t(k − 2) + r + 1− d

= p− 1 +n(k − 2)

k − 1+

r + 1

k − 1− d = p− 1− p

k − 1+

r + 1

k − 1

• If p ≥ 2, then p− 1− pk−1

≥ p− 1− p2

= p−22≥ 0, the sum is positive.

• If p = 1, then the sum above may be 0 when r = 0 and H = Tk−1,n−1, leading to

n = t(k − 1) + 1 and d = n(k−2)+1k−1

= t(k − 2) + 1. Since n > k we see that t ≥ 2 and sod < n− 1.

In H every color is of t vertices and every vertex is of t(k − 2) degree. After addingx and restoring the deleted edges to make G, there is still at least one vertex of thatdegree, which is smaller than that of x, contradicts the supposition about x.Case 2. |V (Gi)| ≥ k − 1 for i = 1, ..., n− d. So, d ≥ (k − 1)(n− d)Let

d = s(k − 1) + q, 0 ≤ q ≤ k − 2. (1.2)

6

Then 1 ≤ n− d ≤ s ≤ t by 1.2. The minimum degree in G(N(x)) is at least

d− (n− d) ≥ d− s

and so, according to T 3, there exist vertex disjoint complete subgraphs G1, G2, ..., Gs ofk − 1 vertices in G(N(x)) so that ∪s

1V (Gi) ⊆ V (N(x)).Again, using the induction hypothesis for the graph

H := (V (G)− x,E(G)\ ∪s1 E(Gi) ∪ E(x))

of tk−1,n + m− d− s(

k−12

)edges, we have

edkk(G) ≥ s + edkk(H)

≥ s +1(

k2

)− (k − 2)(tk−1,n + m− d− s

(k − 1

2

)) =

1(k2

)− (k − 2)m

+1(

k2

)− (k − 2)(s + t(k − 2) + r − d) >

1(k2

)− (k − 2)m.

Heres + t(k − 2) + r − d = s + t(k − 2) + r − s(k − 1)− q

= t(k − 2)− s(k − 2) + r − q = (t− s)(k − 2)− q + r > 0

Let’s explain why this happens.

• if t = s, then the sum is not negative, and could be zero if q = r, that is d = n− 1, andH = Tk−1,n−1. However, in H, each color does not contain edges because we have deletedfrom them, but we only deleted edges that form edge disjoint Kk−1 in G−x. So each coloris of k − 1 size, it means n = (k − 1)2 + 1. It is easy to verify that in this graph, a copyof K(k−1)2+1) where k ≥ 4, there are at least k, and then more than k − 1 = 1

(k2)−(k−2)

m

edge disjoint K ′ks.

• if t > s then q ≤ k−2, the sum is not negative. It would be 0 if t = s+1, q = k−2, r = 0,and H = Tk−1,n−1, so d = (t − 1)(k − 1) + k − 2 = t(k − 1) − 1 = n − 2. But the in H,again, every vertex is of t(k−2) degree. After adding x and the deleted edges to make G,since there is one vertex is not x′s neighbor, its degree remains unchanged, in addition,t(k − 1)− 1 > t(k − 2), so x could not be a vertex of minimum degree in G

The result of Gyori and Tuza is proved for k ≥ 4 already. As we have noticed, theK6m−2 example disproves the similar assertion for k = 3: p3(G) ≤ 2t2,n. However, itmight be possible that:

7

Conjecture. For every graph of n vertices, p3(n) ≤ 12n2 + o(n2).

(⇔ edk3(t2(n) + m) ≥ 23m− o(n2)).

Gyori and Tuza [6] proved

T 4. p3(n) ≤ 916

n2.

We prove the following improvement

Theorem 1.2. edk3(n, t2(n) + m) ≥ 59m−O(n).

Before proving the theorem, we need

T 5. Let k3(G) denote the number of triangles in an arbitrary graph G of n vertices. Then

G contains at least k3(G)n−2

−O(n) edge disjoint triangles.

Proof of T 5. It has been shown proven by Spencer [10] that in the complete graph Kn

of n vertices there can be found a (partial) Steiner system Sn of sn = 13

(n2

) − O(n) edgedisjoint triangles.Let G be a graph of n vertices and denote by T the set of triangles of G(then|T| = k3(G)).For an arbitrary permutation π of V (G), put tπ = |T ∩ π(Sn)| where π(Sn) is the imageof Sn after applying π.For each pair T, T ′ of triangles satisfying T ∈ T and T ′ ⊂ Kn there are exactly 6(n− 3)!permutations such that T = π(T ′). On the other hand, Sn contains sn of the triangles ofKn, therefore the average value of tπ is k3(G)sn/

(n3

). Consequently, there exists a π for

which π(Sn) contains at least k3(G)sn/(

n3

)triangles of G.

T 6. Every graph of n vertices and e edges contains at least e3n

(4e−n2)( 1n−2

)−O(n) edgedisjoint triangles.

Proof of T 6. Without loss of generality, we replace t2(n) by n2

4.

If e ≤ n2

4, there is nothing to prove.

Otherwise, if n2

4< e ≤ (

n2

), the result of Moon and Moser [11] states that G contains at

leastk3 ≥ e

3n(4e− n2)

triangles, therefore T 5 implies the validity of our theorem.

Proof of Theorem 1.2. Induction on n.The statement is obvious for n ≤ 5.Assume that Theorem 1.2 holds for graphs of at most n− 1 vertices.Let G be an arbitrary graph of order n and size e = n2

4+ m

Case 1. e ≥ 5n2

12then T 6 implies

edk3(G) ≥ e

3n2(4e− n2)−O(n)

8

≥512

n2

3n24m−O(n) =

5

9m−O(n)

Case 2. e < 5n2

12.

Denote by d the minimum degree, then d < 56n. Let x be a vertex of G with this minimum

degree, N(x) be the graph induced by x′s neighbors. We distinguish two subcases.

Subcase 2.1 If d ≤ n2

then

edk3(G) ≥ edk3(G− x) ≥ 5

9(e− d− (n− 1)2

4)−O(n− 1)

=5

9(e− n2

4) +

5

18(n− 2d)−O(n) ≥ 5

9m−O(n).

Subcase 2.2 If d > n2, in N(x) there exist vertex disjoint complete subgraphs G1,...,Gn−d

of order b dn−d

c or d dn−d

e satysfying ∪n−d1 V (Gi) = V (N(x)).

Suppose that |V (Gi)| = d dn−d

e for i = 1,..,p and |V (Gi)| = b dn−d

c for i = p + 1, ..., n − dwhere p ≤ n− d− 1If b d

n−dc = 1, then n

2< d < 2n

3and p = 2d− n.

We see that

edk3(G) ≥ p + edk3(G− x, e− d− p) ≥ p +5

9(e− d− p− (n− 1)2

4)−O(n− 1)

=5

9(e− n2

4) + p +

5

18n− 5

9(d + p)−O(n)

=5

9m−O(n) +

6d− 3n

18≥ 5

9m−O(n).

If b dn−d

c ≥ 2 then letd := 2s + q(q = 0, 1), (1.3)

In G(N(x)) every degree is at least d−s(2d−n ≥ d−s), so there exist s endpoint disjointedges. Therefore

edk3(G) ≥ s + edk3(n− 1, e− d− s)

≥ s +5

9(e− 3s− q − (n− 1)2

4)−O(n− 1)

=5

9(e− n2

4)−O(n) +

5

18n− 2

3s

≥ 5

9m−O(n)

since 2s ≤ d < 56n. We are done.

9

Corollary. p3(n) ≤ 59n2 + O(n).

Proof of corollary. Obviously, p3(G) = 2e− 3edk3(G). Using both

edk3(G) ≥ 4m(m− n2

4)

3n2

and

edk3(G) ≥ 5

9m−O(n)

for every graph G of size e more than n2

4we obtain

p3(G) ≤ n2 m

n2(3− 4

m

n2)

and also

≤ m

3+

5

12n2 + O(n)

either. The maximum point of the smaller value between them does not exceed 59n2+O(n),

being when mn2 is about 3

8

I believe that by induction, as well as by combination of estimates of edk3(G) depend-ing on the magnitude of e and δ(G), we might work out better results.

For k = 4, Theorem 1.1 gives us edk4(n, t3(n)+m) ≥ 14m. Using the analogous technique,

this result can be refined.

Theorem 1.3.

edk4(n,m + t3(n)) ≥ 5

18m− o(n2).

To settle this theorem, a general result of Moon and Moser dealing the number ofcliques is needful. Let t(Kr, G) = hom(Kr,G)

nr where hom(Kr, G) denote the number of

homomorphism of Kr to G, that is t(Kr, G) = r!kr(G)nr .

T 7. r t(Kr,G)t(Kr−1,G)

≤ (r − 1) t(Kr+1,G)t(Kr,G)

+ 1.

Proof of T 7. H ir+1 denotes the graph remaining after deleting i edges sharing a common

vertex in Kr+1. It can be shown that t(Kr−1, G)t(H1r+1, G) ≥ t(Kr, G)2 (from the positive

semidefinite property of a matrix, [13])Consequently,

rt(Kr, G)

t(Kr−1, G)≤ r

t(Hr+1, G)

t(Kr, H).

Applying the super-modular property(i.e t(F+e,G)+t(F+f, G) ≤ t(F, G)+t(F+e+f, G))r − 1-times, we obtain that rt(H1

r+1) ≤ (r − 1)t(Kr+1) + t(Kr).

10

Corollary. for k ≥ 2

kk(G) ≥ 1

k!

k−1∏1

(1− i(1− 2q))nk where q =e

n2. (1.4)

Proof of corollary. There is nothing to prove if e < 12(1 − 1

k−1)n2. Suppose that e ≥

12(1− 1

k−1)n2. Let bi+1 := ti+1(G)

ti(G), i = 1, .., k − 1, b1 := 1. Note that bi ≤ 1(i = 1, .., k) and

tk(G) =∏k

1 bi.The inequality 1.4 implies

rbr ≤ (r − 1)br+1 + 1, so r(1− br) ≥ (r − 1)(1− br+1).

therefore

1− bi ≤∏i−2

1r+1

r(1− b2) = (i− 1)(1− 2e

n2 ) = (i− 1)(1− 2q), bi ≥ 1− (i− 1)(1− 2q).

Corollary.

k4(G) ≥ 1

6q(3q − 1)(4q − 1)n4 where q =

e

n2≥ 1

3. (1.5)

We still need the general form of the T 5

T 8. G contains at least (r−2)!kr(G)nr−2 − o(n2) edge disjoint Kr.

Proof of this theorem. According to a result of V. Rodl [18], There is a system of edgedisjoint K ′

rs which covers almost all the edges of the complete graph Kn but o(n2). Denotethis system by Sn, the left steps are quite similar to what we have done in T 5

Proof of Theorem 2.2. We will use the same technique as before. For convenience, lett3(n) be n2

3. To explain why the constant is 5

18, let us replace it by α and try to find the

possibly largest value of α.For small n, clearly edk4(G) ≥ αm − o(n2). Suppose that it holds for every graph of atmost n− 1 vertices. Consider an arbitrary graph G of n vertices and let x be a vertex ofG with minimum degree, d(x) = d.If d is relatively small then the induction step can be done easily as follows. In case ofd < 2n

3, we have

edk4(G) ≥ edk4(G− x) ≥ α(e− d− (n− 1)2

3)− o((n− 1)2)

≥ αm− o(n2).

11

It is still easy when 2n3≤ d < 3n

4because applying the T 3 to G(N(x)), there exist vertex

disjoint subgraphs G1,...,Gn−d of order 2 or 3 which cover the vertices of G(N(x)). Thenumber of K ′

3s in {G1,...,Gn−d} is 3d− 2n, therefore

edk4(G) ≥ p + edk4(G− x) ≥ p + α(e− d− 3p− (n− 1)2

3)− o(n2)

= α(e− n2

3) + (3− 10α)(d− 2n

3)− o(n2) ≥ αm

Providing that

α ≤ 3

10.

So by this way we can’t hope to find α larger than 310

From now on let d ≥ 3n4

.Set d = 3s + q, where 0 ≤ q ≤ 2. Again, since every vertex in G(N(x)) has a degree atleast d− s, there exist s vertex disjoint K3 in G(N(x)), as a consequence

edk4(G) ≥ s + edk4(n− 1, e− d− 3s))

≥ s + α(e− d− 3s− (n− 1)2

3))− o((n− 1)2)

≥ αm− o(n2) +2α

3(n− d(6α− 1)

2α).

Thus if e ≤ α6α−1

n2 then we accept this induction step. Otherwise the formula stated inT 8 for r = 4 gives,

edk4(G) ≥ 1

3q(3q − 1)(4q − 1)n2 − o(n2),

here q = en2 ≥ α

6α−1and q ≥ 1

3also.

Under these conditions, the inequality

13q(3q − 1)(4q − 1) ≥ α(q − 1

3) holds when α ≤ 5

18.

A possible generalization of these theorems is

Conjecture. edkk(n, tk−1(n) + m) ≥ 2km− o(n2).

Remarks.Let p(G) denote the minimum of

∑ |V (Gi)| over all decompositions of G into edge disjointcliques G1, G2, .. (i.e. the subgraphs Gi’s are pairwise edge disjoint cliques such thatE(G) = ∪E(Gi) and they are not of given sizes). It was conjectured by G.O.H. Katonaand T. Tarjan and was proven by F.R.K. Chung, E. Gyori and A.V. Kostochka that

12

T 9.p(G) ≤ 2t2(n)

equality holds if and only if G ' T2(n).

The proof is now simple. So if G is K4 free then edk3(G) ≥ 23m, It is still easy to show

that in this case edk3(G) ≥ 34m

P. Erdos suggested to study a similar weigh-function. For any graph G, let p∗(G) denotethe minimum of

∑(|V (Gi)| − 1) over all decompositions of G into pairwise edge disjoint

cliques G1, G2, .... A maybe too optimistic conjecture is:

Conjecture. For every graph G of n vertices,

p∗(G) ≤ t2(n).

This conjecture seems to be just a bit stronger than the T 9, but it is not the case. Aspecial case of it is still open, as well.

Conjecture. Every K4-free graph of n vertices and t2(n) + m edges contains m edgedisjoint triangles.

Only the following even weaker special case is settled by Gyori.

T 10. Every 3-colorable graph of n vertices and t2(n) + m edges contains m edge disjointtriangles.

Proof of T 10. Suppose A,B, C with cardinals a ≥ b ≥ c be the three color parts ofV (G). For M is an arbitrary subset of A with cardinals b let edk3(M, B, C) denote themaximum edge disjoint triangles of G restricted on M, B, C. Then edk3(M,B, C) ≥bc − nBC − nBM − nCM where nXY := |X||Y | − eG(X, Y )(the number of not-G graphedges between X and Y )So the average

∑M⊂A edk3(M,B, C)(

ab

) ≥ bc− nBC −nAC

(a−1b−1

)+ nAB

(a−1b−1

)(

ab

)

= bc− nBC − b

a(nAC + nAB) ≥ bc− (nBC + nAC + nAB)

= bc− (ab + bc + ca−m− t2(n)) = m + t2(n)− a(b + c) ≥ m.

13

To end up the chapter, we examine the graph G of order n, formed by adding m edgesto the T2,n.Let m1,m2 be the number of added edges in the red, as well as in the blue color of G, letci = mi

n2 , c := c1 + c2 = mn2 . We estimate the number of edge disjoint triangles as follows.

Suppose there are m1 not monochromatic edge disjoint triangles that cover the m1 rededges and m2 not monochromatic edge disjoint triangles that cover the m2 blue edges. Bypermuting the vertices in each color we receive many such covering by triangles. Let uschoose one of those examples randomly and consider the triangles as vertices, adjancencymeans that the triangles share some edge. We are finding maximal independent set ofvertices. Note that a graph of n edges and e edges have at least n− e independent set ofvertices. By counting the expectation of the size of the independent set, since each pairof red-blue edges add 4 to the amount, it is at least

m1 + m2 − 4m1m2

n2

2 ≥ m− 4m2

n2

As we can see, when m = o(n2), the estimate works and almost the best, results in aboutm. However, when m = O(n2) it is unusable because we just used the not monochromatictriangles, for example if m ≥ 1

8n2 then we win not more than m

2triangles.

It is likely that there are at least 23m edge disjoint triangles in G(would be a corollary of

the first conjecture.) However I have no idea to verify this fact. A special case, when the medges form a triangle-free subgraph in the two colors (therefore only not-monochromatictriangles exist) might be examined by using the fractional results discussed in Chapter 3.

Similarly, the result can be verified for k ≥ 3, that is there exists a constant ck suchthat after adding m edges to Tk−1,n we must have m− ck

m2

n2 edge disjoint K ′ks.

14

Chapter 2

If we add m edges to the Turan graph Tk−1(n) arbitrarily then the resulting graph containsat most m edge disjoint cliques of k vertices because each clique must contain at least onenew edge. that is,

edkk(n, tk−1(n) + m) ≤ m

If m is large enough, it may occur that we can’t find m or asymptotically m edge disjointcliques of k vertices.To examine the case when m is small, first we need the following([12])

Let k be an integer ≥ 3, and G be a graph of order n so that tk−1(n) ≤ |E(G)|. Wewrite |E(G)| in the form

|E(G)| = (1− 1

t)n2

2Let d be btc, set m = |E(G)| − td(n) then

T 11. (“stability theorem”)For C be an arbitrary constant there exist positive constants ∆ = ∆(C) and D = D(C)such that if 0 < m < ∆n2 and

kk(G) ≤(

t

k

)(n

t)k + Cmnk−2

then there exists a Kd(n1, .., nd) such that∑

ni = n, |ni − nd| < D

√m, and G can be

obtained by adding to and deleting from this Kd(n1, n2, .., nd) less than Dm edges.

Especially, for k = d + 1, suppose that m = E(G)− tk−1(n) is much smaller than n2,the theorem becomes

Corollary If kk(G) ≤ Cmnk−2 then there exists a Kk−1(n1, .., nk−1) such that∑

ni =n, |ni − n

k−1| < D

√m and G can be obtained by adding to and deleting from it less than

Dm edges.

Now we are ready to prove

15

Theorem 2.1. edkk(n, tk−1(n) + m) = m−O(m32

n) if m = o(n2).

Proof of theorem 2.1. Set C be any constant larger than 1.If

kk(G) ≥ Cmnk−2

then by the proof used in T.8, G contains

kk(G)k!(n− k)!((n

2)(k2)− o(n2))

n!≥ kk(G)n2(k − 2)!(1− o(1))

nk

= Cm(k − 2)!(1− o(1))

edge disjoint copies of Kk, which is more than m provided that n is sufficiently large, andwe are done.Otherwise, by the corollary,

kk(G) ≤ Cmnk−2

implies there exists Kk−1(n1, .., nk−1) such that∑

ni = n, |ni − nk−1| < D

√m and G can

be obtained by adding to and deleting from it less than Dm edges. Suppose that thenumber of deleted edges is mdel. Let mnom be the number of monochromatic G-edges,then

O(m) = mnom ≥ mdel + m. (2.1)

Set mi be the number of monochromatic edges in i-th color. In each color we choosen

k−1−D

√m vertices such that the number of monochromatic edges in it maximum, that

is it contains at least

(n

k−1−D

√m

nk−1

+ D√

m)2mi

chromatic edges. Therefore the chosen vertices form a Turan graph(with colors’ of sizen

k−1−D

√m) but at most mdel edges have been deleted and the number of the remaining

monochromatic edges, m′nom, is obviously

≥ (n

k−1−D

√m

nk−1

+ D√

m)2mnom = mnom −O(

m32

n). (2.2)

According to the result noted at the end of the previous chapter, there are at least

m′nom −O(

m′nom

nk−1

−D√

m)2 −mdel

edge disjoint copies of Kk, Taking account into 2.1, 2.2 we see that

edkk(G) ≥ m−O(m

32

n)

16

For the triangle-case(and similarly for k ≥ 4), let us produce a direct proof of

edk3(n, t2(n) + m) = m−O(m

32

n)

if m = o(n2).

Proof of . Roughly speaking, we will prove that either we can find m edge disjoint trian-gles or the graph is similar to a graph obtained from T2(n) by adding m edges to it.Delete the edges of edge disjoint triangles of maximum number and let G∗ denote theresulting triangle-free graph. If we found m edge disjoint triangles then we are done. Sowe assume that

|E(G∗)| > t2(n)− 2m

Choose a maximum size spanning bipartite graph G0 of G∗(with two coloring A,B), letG1 = (V,E(G∗) − E(G0)), i.e the spanning subgraph of the monochromatic edges. Notethat dG0(v) ≥ dG1(v) ( otherwise by changing the color of v we would obtain a greatersize spanning bipartite subgraph)Proposition

E(G1) = O(m

32

n)

ProofSince

1

|E(G∗)|∑

uv∈E(G∗)

(d∗G(u) + d∗G(v)) =1

|E(G∗)|∑v∈V

(d∗G(v))2

≥ n

|E(G∗)|(∑

v∈V d∗G(v)

n)2

=4|E(G∗)|

n≥ n− 8m

n.

There is an edge uv ∈ E(G∗) such that d∗G(u) + d∗G(v) ≥ n− 8mn

.Let H0 be a spanning bipartite graph of G∗ whichi) contains the edges incident to u or v.ii) dH0(w) ≥ dH1(w) if w is not u nor v neighbor, H1 = (V, E(G∗)− E(H0)).There exists such H0, for example a maximum size spanning bipartite subgraph satisfyingi). We have

E(H1) ≤ d8m

n

Where d is the maximum degree in H1 because the vertices that are not adjacent to u orv represent all the edges of H1.

17

If x is a vertex that dH1(x) = d then the graph G∗ does not contain any edge joiningNH0(x) and NH1(x), so

t2(n)− 2m ≤ |E(G∗)| = |E(H0)|+ |E(H1)|

≤ t2(n)− dH1(x)dH0(x) + d8m

n

≤ t2(n)− d2 + d8m

n

Thus,

d ≤ 4m

n+

√16m2

n2+ 2m ≤ (

√2 + o(1))

√(m)

and

|E(H1) = O(m

32

n).

which implies that E(G1) = O(m32

n) by the choice of G0.

The proposition implies that

|E(G0)| ≥ t2(n)− 2m−O(m

32

n) = t2(n)− (2 + o(1))m

and|A|, |B| ≥ n

2−O(

√m).

From here we may proceed as above.

However, by more careful techniques, we can also prove that

Theorem 2.2. If m = |E(G)| − tk−1(n) = o(n2) then either G contains m edge dis-joint K ′

ks or there is an (k − 1)-coloring {V1, ..Vk−1} of V (G) such that the number of

monochromatic edges is tk−1(n)−O(1)m2

n2 and so |Vi| = nr−1

+ O(1)mn

for i = 1, .., k − 1.

Using this theorem, it is easy to prove that

Theorem 2.3. (Gyori [8])

edkk(n, tk−1(n) + m) = m−O(m2

n2) if m = o(n2)

18

Corollary. There is a function f(c)(= c0c2) such that every graph of n vertices and

tk−1(n) + cn edges contains cn− f(c) edge disjoint K ′ks.

It can be proved that for a given order n and size tk−1(n) + m (m = o(n2)), edkk(G)is minimum if G is obtained from Tk−1(n) by adding edges so that bm

nc vertices of each

color join to all vertices.It impliesT 12. There is a constant c that edk3(t2(n) + m) ≤ m− cm2

n2 if 2n ≤ m = o(n2).(and we

can do also for k ≥ 4, so the O(m2

n2 ) factor is tight in the 2.3.)Proof. It is enough to estimate the edk3(G) where G is obtained from T2(2k) by addingm = an − a(a + 1), 2 ≤ a = o(n) edges so that a vertices of each color should be joinedto all other vertices. Let the two colors be U, V and AU , AV be the two set of a vertices.For a set S of edge disjoint triangles let G0 be the graph containing monochromatic edgesthat belong to the not chromatic triangles of the set.Let m0,m1 be the number of not chromatic and chromatic triangles of the set.The chromatic triangles contain 3, the not chromatic ones contain 1 new edges, so

m0 + 3m1 ≤ m

It implies that

|S| = m0 + m1 ≤ m

3+

2m0

3Obviously

m0 =∑

ui∈AUdG0(ui) +

∑vi∈AV

dG0(vi)− |G0-edges having both endpoints in AU or AV |

For u ∈ AU we have taken at least 2(dG0(u) + a− k) not monochromatic edges that havean endpoint in AV to cover dG0(u) not monochromatic triangles at u, the sum using tocover all can be at least(each edge can be counted twice)

∑ui∈AU

(dG0(ui) + a− k) =∑

ui∈AU

dG0(ui) + a2 − ak

On the other hand we have to use at least∑

vi∈AVdG0(vi) not monochromatic edges to

cover AV−vertices, hence∑

vi∈AV

dG0(vi) ≤ ak − (∑

ui∈AU

dG0(ui) + a2 − ak)

that is ∑ui∈AU

dG0(ui) +∑

vi∈AV

dG0(vi) ≤ 2ak − a2 = m + a

19

Therefore if the number of |G0-edges having both endpoints in AU or AV is at least a2

2

then m0 ≤ m + a− a2

2.

Otherwise, of less than a2

2edges in AU and AV are used in the not monochromatic triangles.

Thus m0 ≤ m− a2

2.

So

edk3(G) ≤ m− a2

3+ O(a) = m−O(

m2

n2).¤

Another relating problem, questioned by P.Erdos, is

Problem. Determine the maximum m such that every graph of n vertices and t2(n)+medges has m edge disjoint triagles. The answer follows

Theorem 2.4. (Gyori [7])

edk3(n, t2(n) + m) = mif m ≤ 2n− 10 for odd nif m ≤ 3

2n− 5 for even n

providing that n is sufficiently large.

Examples show that the upper bounds for m in Theorem 2.4 are sharp.

Example 1. n = 2kAdd 3k − 4 edges to T2(n) so that two vertices of color class A and one vertex of colorclass B should be joined to all vertices. If the k− 1 edges in B are covered by k− 1 edgedisjoint triangles then deleting these 3k − 3 edges, every vertex in A has k − 2 neighborsin B except one which has k ones. Since two vertices in A have k − 1 neighbors in A, wecan’t find 2k − 3 edges disjoint triangles covering the 2k − 3 edges incident to these twovertices.This is the unique extremal example.

Example 2. n = 2k + 1Take the Turan graph T2(n) with color classes V1 and V2 of k and k + 1 vertices, respec-tively. Take one vertex in V1 and join it to all vertices in V1. Then, take three vertices,say x1,x2,x3 in V2, and join them to each other and to k − 3(except one) other verticesin V2.(The set of k − 3 vertices do not have to be the same for the vertices x1, x2, x3) Itis easy to see that the resulting graph G of n vertices and t2(n) + 2n− 9 edges does notcontain 2n− 9 edge disjoint triangles.We have only three extremal examples.

The continuing of 2.4 follows

20

Theorem 2.5. (Gyori [8])

edkk(n, tk−1(n) + m) = m if m ≤ b3n + 1

r − 1c − 5 for r ≥ 4

provided that n is sufficently large.

The following example shows that the upper bound for m in Theorem 2.5 is sharp.

Example 3.Consider two color classes Vi and Vj of T(k − 1)(n) of bn+1

r−1c vertices. Add 3bn+1

r−1c - 4

edges to T(k − 1)(n) so that two elements of Vi and one element of Vj should be joinedto all vertices. If the bn+1

r−1c − 1 edges in Vj are covered by bn+1

r−1c − 1 edge disjoint K ′

ks

then deleting these (bn+1r−1c− 1)

(n2

)edges, every vertex in Vi has bn+1

r−1c− 2 neighbors in Vj

except one which has bn+1r−1c neighbors. Since two elements of Vi have bn+1

r−1c− 1 neighbors

in Vi, we cannot find 2bn+1r−1c− 3 edge disjoint K ′

ks covering the 2bn+1r−1c− 3 edges incident

to these two vertices.This is the only extremal graph.

RemarksAnother important characterization of the Turan graphs isLet Gk−1,m(n) denote a graph obtained from Tk−1(n) by adding m edges so that the newedges belong to the same class having maximum number of vertices(i.e. d n

k−1e) and the

new edges do not form triangles, if this is possible. Then there exist a constant ck > 0such that for m < ckn, if G is a graph of order n and size tk−1(n) + m then

kk(G) ≥ kk(Gk−1,m(n)) = m∏

0≤i≤k−3

bn + i

k − 1c (2.3)

Problem Determine the constant ck in the theorem above.

If we add p+1 or more edges to the first class of Kk−1(p+1, p, .., p, p− 1), then each newedge will be contained only in (p− 1)pk−3 K ′

ks, and we can verify that 2.3 does not hold.So ck ≤ 1

k−1

Theorem 2.6. [12] If e(G) = tk−1(n) + m, where m < b nk−1c, then

kk(G) ≥ m∏

0≤i≤k−3

bn + i

k − 1c

For example, k = 3 gives

21

If e(G) = t2(n) + l, 0 < l ≤ n2, then G contains at least lbn

2c triangles.

Here we reproduce a simple fact proving for the case l ≤ n4, as follows[3]

We assume that n is even(similar of the case n odd)Let tk (k=1,2,3) be the number of triangles in Kn that contain exactly k edges of G. Thust0 = k3(G).Let λv be the number of edges between N(v) and V (G − N(v). For uv edge, denote bytkuv the number of triangles in Kn that contain uv and k edges of G. By definitions, it isobviously that ∑

v

λv = 2t2, (2.4)

∑

uv∈E(G)

t0uv = 3t0, (2.5)

∑

uv∈E(G)

t2uv = t2, (2.6)

n = d(u) + d(v)− t0uv + t2uv, (2.7)

provided that uv ∈ E(G).Summing the last equality over all edges uv, and by 2.5, 2.6, we have

nm =∑

u

d(u)2 − 3t0 + t2, (2.8)

Since ∑

uv∈E(G)

(d(u) + d(v)) =∑

u

d(u)2.

By putting p(u) = d(u)− n2

we obtain that

∑u

d(u)2 =∑

u

p(u)2+n∑

u

p(u)+n3

4=

∑u

p(u)2+sn(m−n2

4)+

n3

4=

∑u

p(u)2+2ln+n3

4.

Together with 2.4 and 2.8, it gives

3t0 =∑

u

(p(u)2 +λ(u)

2) + ln.

This relation gives the assertion of the theorem unless

∑u

(p(u)2) +λ(u)

2≤ ln

2. (2.9)

22

Thus, we may assume 2.9 holds. Then for some vertex u

p(u)2 +λ(u)

2≤ l

2. (2.10)

The number of edges in G with both end vertices in N(u) or V (G)−N(u) is

n2

4+ l − (

n2

4− p(u)2 − λ(u)) = l + p(u)2 + λ(u).

Each such edge belongs to at least

n

2− |p(u)| − λ(u) ≥ n

2− p(u)2 − λ(u)

G-triangles (which have the third vertex in the other color). Therefore

k3(G) > ((l + p(u)2) + λ(u))(n

2− p(u)2 − λ(u)) =

ln

2+ (p(u)2 + λ(u))(

n

2− l − p(u)2 − λ(u))

≥ ln

2+ (p(u)2 + λ(u))(

n

2− 2l) >

ln

2¤

23

Chapter 3

Part 1.In the beginning of Chapter 1 we have defined:

pk(G) =min{∑m1 |V (Gi)| : Gi’s are Kk and edges,∪m

i E(Gi) = E(G)}.

The weight of each K ′ks copies is k and 2 is that of edges. We also proved that

pk(G) ≤ 2tk−1,n

for k ≥ 4.In the first part of this chapter, we use the fractional method from [5]. We will study theproblem when the weight of K ′

ks copies is a given integer s less than k(k − 1).Let’s start with some definitions.For G is a graph of n vertices, a ψ∗ function on subgraphs H of G where H ' Kk or anedge to [0, 1] is called fractional full-packing if

∑e∈H

ψ∗(H) = 1

for every edge e.

Definition. |ψ∗s | :=∑

H'Kksψ∗(H) +

∑H'K2

2ψ∗(H).

Definition. ψ∗s(G) := min{|ψ∗s |, ψ∗ is fractional full-packing of G}.

A ψ fractional full-packing to {0, 1} is called integer full-packing.ψs(G) := min{|ψs|, ψ is an integer full-packing of G}.It is obvious that ψs(G) ≥ ψ∗s(G), the following shows that the difference is small

Theorem 3.1.ψs(G)− ψ∗s(G) = o(n2)

for all graphs G.

24

The theorem in fact is a direct corollary of a deep application of the Szemeredi regu-larity lemma [19], as followsLet H0 be any fixed graph, let νH0(G) denote the maximum size of a set of pairwise edgedisjoint copies of H0 in G and ν−H0

(G) denote the maximum value of∑

H∈G,H'H0ψ(H),

where ψ runs over all functions on copies of H satisfying∑

e∈H ψ(H) ≤ 1 for every edgee of G(i.e fractional packing). Then

ν−H0(G)− νH0(G) = o(|V (G)|2)

for all graphs G.This theorem makes it easier to study the edkH0(G) when the structure of the graph isregular.

Proof of Theorem 3.1. We see thatψ∗s(G) = min{∑H'Kk

sψ∗(H) + 2∑

H'K2ψ∗(H) : ψ∗ is a fractional

full-packing of G}= min{2 ∑e

∑H'Kk or edge,e∈H ψ∗(H)− (k(k − 1)− s)

∑H'Kk

ψ∗(H) :

ψ∗ is a fractional full-packing of G}= 2e−(k(k−1)−s)ν−Kk(G), since max

∑H'Kk

ψ∗(H) =

ν−Kk(G).

Similarly, ψs(G) = min{∑H'Kksψ(H) + 2

∑H'K2

ψ(H) : ψ is an integerfull-packing of G}= min{2 ∑

e

∑H'Kk or edge,e∈H ψ(H)− (k(k − 1)− s)

∑H'Kk

ψ(H) :ψ is an integer full-packing of G} = 2e− (k(k − 1)− s)νKk

(G)due to max

∑H'Kk

ψ(H) = νKk(G).

Definition. p∗s(n) := max{|ψ∗s(G)|, G is of order n}, ps(n) := max{|ψs(G)|, G is oforder n}.Corollary. ps(n)− p∗s(n) = o(n2)The pk(n) has been investigated. By the corollary we can study p∗s(n) instead of ps(n).The p∗s(n) can be studied easily.

Theorem 3.2. The p∗s(n)n(n−1)

sequence is monotone decreasing.

Proof of Theorem 3.2. Let G be a graph of order n + 1 where |ψ∗s(G)| = p∗s(n + 1).For every vertex v of G let Gv is the G− v graph, and ψ∗Gv

is its optimal fractional full-packing.Then, the ξ∗ := 1

n−1

∑v ψ∗Gv

is also a fractional full-packing inG.Obviously,

|ξ∗s | =1

n− 1

∑v

|ψ∗Gv| ≤ n + 1

n− 1p∗s(n)

and |ξ∗s | ≥ min{|ψ∗|, ψ∗ is a fractional full-packing of G} = ψ∗s(G) = p∗s(n + 1).

Corollary. The limit of p∗s(n)n2 , and also, of ps(n)

n2 exists when n −→∞, they are equal.

25

We have another relation

Theorem 3.3. p∗s(kn) ≤ k2p∗s(n) + sn + bR(k,k)−1k

c(k2 − k − 2− s)+

where R(k, k) is the appropriate Ramsey number.

Proof of Theorem 3.3. Let Gkn be a graph of order kn where |ψ∗s(Gkn)| = p∗s(kn).We can divide the graph into n groups G1, G2, .., Gn of k vertices so that each of the firstn− bR(k,k)−1

kc groups is Kk or its complement.

For each v1, v2, .., vn vertices from each group, define ψ∗v1,v2,..vnbe an optimal fractional

full-packing of the graph formed by v1, v2, ..vn. Let ψ∗i be an optimal fractional full-packingof the i− th group(graph formed by the vertices in group).Clearly,

ξ∗ :=1

kn−2

∑vi∈Gi

ψ∗v1,v2,..vn+

∑i

ψ∗i

is a fractional full-packing in Gkn, therefore |ξ∗s | ≥ p∗s(kn)On the other hand,

|ξ∗s | ≤1

kn−2knp∗s(n) + (n− bR(k, k)− 1

kc)s + bR(k, k)− 1

kcmax{s, k(k − 1)− 2}

= k2p∗s(n) + sn + bR(k, k)− 1

kc(k2 − k − 2− s)∗,

and we are done.

Let cks be the limit in Theorem 3.2.

In Chapter one we have shown some lower bounds of ckk, not knowing there exist or not.

As k = 3 and s = 3, since R(3, 3) = 6, the Theorem 3.4 shows:

p∗3(3n) ≤ 9p∗3(n) + 3n + 1. (3.1)

The c33 ≤ 5

9is a result in chapter one.

Theorem 3.4. c33 ≤ p∗3(n)

n2 + 12n

+ 18n2 , and hence c3

3 ≤ p3(n)n2 + 1

2n+ 1

8n2 too

Proof of Theorem 3.4. From 3.1,

p∗3(3k+1n) ≤ 9k+1p∗3(n) +

k∑i=0

9k−i(3i+1n + 1).

This implies

c33 ≤

p∗3(3k+1n)

3k+1n(3k+1n− 1)=

3k+1n

3k+1n− 1

p∗3(3k+1n)

9k+1n2

26

≤ 3k+1n

3k+1n− 1(p∗3(n)

n2+

1

3n

k∑i=0

3−i +1

n2

k∑i=0

9−i−1).

Also the limit of the latter term isp∗3(n)

n2 + 12n

+ 18n2 when k tends to infinity

Similarly, the k = 4, s = 4 case(R(4, 4) = 18):

p∗4(4n) ≤ 16p∗4(n) + 4n + 24. (3.2)

and we have

Theorem 3.5. c44 ≤ p∗4(n)

n2 + 13n

+ 110n2 . Consequently, c4

4 ≤ p4(n)n2 + 1

3n+ 1

10n2 too.

Proof of theorem 3.5. From 3.2

p∗4(4k+1n) ≤ 16k+1p∗4(n) +

k∑i=0

16k−i(4i+1n + 24)

Which leads to

c44 ≤

p∗4(4k+1n)

4k+1n(4k+1n− 1)=

4k+1n

4k+1n− 1

p∗4(4k+1n)

16k+1n2

≤ 4k+1n

4k+1n− 1(p∗4(n)

n2+

1

16n

k∑i=0

41−i +24

16n2

k∑i=0

16−i−1).

The limit of the latter term isp∗4(n)

n2 + 13n

+ 110n2 when k tends to infinity

By help of such computation we can determine the value of p3(n)n2 ,p4(n)

n2 ...when n is largeenough, I think the constants 5

9, 5

18... can be improved this way.

Remarks.Let k be unchanged. For x ∈ [0, k(k − 1)], and n is given, let Gx be an optimal graph(orone of),i.e |ψGx| = px(n).We set mx := E(Gx), Tx:= maximal number of edge disjoint K ′

ks in Gx (in fact mx, Tx

depend on Gx), then for any r ≤ s in the [0, k(k − 1)] interval the followings hold(a) Ts ≥ Tr,ms ≥ mr and

(k(k − 1)− r)(Ts − Tr) ≥ 2(ms −mr) ≥ (k(k − 1)− s)(Ts − Tr)

(b) cx is a continuously monotone, convex function.(c)

(s− r)Ts ≥ ps(n)− pr(n) ≥ (s− r)Tr.

27

Proof of (a). By the maximality of ps(n)ps(n) ≥ |ψs(Gr)| = 2mr − (k(k − 1)− s)Tr, but ps(n) = 2ms − (k(k − 1)− s)Ts, So

2(ms −mr) ≥ (k(k − 1)− s)(Ts − Tr)

By the maximality of pr(n)pr(n) ≥ |ψr(Gs)| = 2ms − (k(k − 1)− r)Ts, but pr(n) = 2mr − (k(k − 1)− r)Tr, So

2(ms −mr) ≤ (k(k − 1)− r)(Ts − Tr)

(b) pr(n) ≥ |ψr(Gs)| ≥ rs|ψs(Gs)| = r

sps(n), thus cs ≥ cr ≥ r

scs, and therefore it is

continuous.If t = αr + (1− α)s, we see that

pt(n) = |ψt(Gt)| = 2mt − (k(k − 1)− t)Tt =

α(2mt − (k(k − 1)− r)Tt) + (1− α)2mt − (k(k − 1)− s)Tt ≤ αpr(n) + (1− α)ps(n).

(c) ps(n)− pr(n) = 2ms− (k(k− 1)− s)Ts− 2mr + (k(k− 1)− r)Tr, and we are done by(b).

Clearly c0 = 1− 1k−1

, ck(k−1) = 1

Corollary.cs − cr

s− r≤ 1

k(k − 1).

If the conjecture edkk(G) ≥ 2km− o(n2) is true, then ck(k−2) = 1− 1

k−1(equivalent), which

means the function cx is constant in [0, k(k − 2)] and linear in [k(k − 2), k(k − 1)].

Part 2.In any 2-coloring of Kn, where n > R(k, k), there are monochromatic K ′

ks. We can askabout the minimum number of such monochromatic K ′

ks.Here we will consider the number of edge disjoint monochromatic K ′

ks.Definition. Let N(n, k) be the minimum number of pairwise edge disjoint monochro-matic complete subgraphs Kk in any 2-coloring of the edges of Kn.

Theorem 3.6. limn→∞N(n,k)n(n−1)

exists .

Proof of theorem 3.6.1. If ck = supn

N(n,k)n(n−1)

, then for any given small ε > 0, there is an m = m(ε) such that

N(m, k)

m(m− 1)≥ (1− ε)ckm(m− 1).

28

Then for n sufficiently large, the edges of Kn can be packed with (1− ε)(n

2)(m

2 )edge disjoint

K ′ms by [18]. As a consequence, for n large

N(n, k) ≥ (1− ε)

(n2

)(

m2

)(1− ε)ckm(m− 1) ≥ (1− ε)2ckn(n− 1).

Therefore we are done, and the limit equals supnN(n,k)n(n−1)

2. Let N−(n, k) be the minimum possible value of ν−Kk(G) + ν−Kk

(G) over all 2-colorings

G ∪G of the edges of Kn.According to a result in Part 1.,

N−(n, k)−N(n, k) = o(n2).

The N−(n, k)′s have the following property

The sequence N−(n,k)n(n−1)

is increasing in n.

Because considering 2-coloring the edges of complete graph Kn+1, for every 1 ≤ i ≤ n+ 1we can find a fractional packing wi of monochromatic Kk of the complete graph receivedfrom Kn+1 after deleting it’s i-th vertex; the total value of wi is at least N−(n, k) bydefinition.Each edge of Kn+1 belongs to n − 1 such complete subgraphs, so the w := 1

n−1

∑i wi is

also a fractional packing for Kn+1. But its total value is at least n+1n−1

N−(n, k). Therefore

N−(n + 1, k) ≥ n + 1

n− 1N−(n, k).

The N−(n,k)n(n−1)

is bounded by 2k(k−1)

for every n. Therefore by the above, this sequenceconverges.

Let ck be this constant; The main problems are to estimate ck for every k.For the case k = 3, if we consider the 2-coloring of Kn determined by T2(n) and itscomplement(two monochromatic cliques) then there are approximately

2

(n/22

)

3=

n2

12+ o(n2)

edge disjoint monochromatic triangles, thus c3 ≤ 112

. This example led to a conjecture ofP. Erdos.Conjecture:

c3 =1

12.

29

The value of N(11, 3) is 6. Therefore, by the first method above,

c3 = supnN(n, 3)

n(n− 1)≥ N(11, 3)

11.10=

3

55.

Another way to find estimate for c3 is by second method and techniques used in the firstpart.

Theorem 3.7. ([5]) N−(3n, 3) ≥ 9N−(n, 3) + n− 1.

Proof of theorem 3.7. Consider such a 2-coloring of K3n that the total value of its maximalpacking is N−(3n, 3). Since 6 vertices contain a monochromatic triangle we can find n−1vertex disjoint monochromatic triangles. Let T1, T2, .., Tn−1 denote these triples and Tn

be the remaining three vertices. For any 3n distinct copies of Kn(from each set eachvertex) there is a fractional packing wi, and because each edge connecting different T ′

js

is covered exactly 3n−2 times by such w′is, the w := 3−(n−2)σiwi is a fractional packing of

the connecting edges. By adding T ′is to this packing, we receive a packing of K3n, whose

value is at least

3−(n−2).3nN−(n, 3) + (n− 1) = 9N−(n, 3) + n− 1

Iterating this result will lead toCorollary

c3 ≥ N−(n, 3)

n2+

1

6n− 1

8n2

for every n ≥ 1The corollary provides that, with the easy fact N−(7, 3) ≥ N(7, 3) = 2,

c3 ≥ N−(7, 3)

72+

1

6.7− 1

8.72=

73

1176>

1

16.11>

3

55

The up-to-now best result of c3 is due to Sudakov and Keevash: c3 > 112.88

Similarly, we might study the case k ≥ 4. It seems that these bounds will depend on theRamsey numbers R(k, k)′s.

RemarkThe conjecture may be weakened by asking whether there are approximately 1

12n2 trian-

gles in the G graph whose complement is triangle-free.We may ask the related question, let N ′(n, k) be the minimum of the maximum of thetwo monochromatic edge disjoint K ′

ks number in any 2-coloring of the edges of Kn. Theconstruction based on the blowup of C5 shows that the number of monochromatic edgedisjoint triangles in either of the colors is about n2

20, and this led to the conjecture of

30

JacobsonConjecture If n is sufficiently large, then

N ′(n, 3) =n2

20+ o(n2).

Obviously, N ′(n, k) ≥ 12N(n, k).

Part 3.In this part, we are discussing on the relations between the graph’s structures.For F is a fixed graph, let ∆F (G) be the minimum number of copies of F and edges whichfully pack E(G), ∆+

F (G) be the minimum number of copies of F and edges which coverE(G) and finally, µF (G) be the maximum number of edges in the F -free subgraphs of G.Clearly,

∆F (G) ≥ ∆+F (G)

According to [16], ∆Kk(G) ≤ tk−1,n for any graph G of order n

for F = K3, we have

Theorem 3.8. ([9]) ∆+K3

(G) ≤ µK3(G)

Shortly, for every e edge ∈ E −M(M is a maximal triangle free subgraph) we find anf edge ∈ M so that e and f are edges of a triangle, and no e′s have the same f . The Hallcondition can be verified by using the maximality of M and the fact that every graphcontains a triangle free subgraph of size at least it’s one half.More generally, the theorem is true for all F has the chromatic number not less than3(one can check that for some bipartite F ′s the theorem does not stand, on the otherhand, for these F the size of a F -free subgraph always be o(n2)), and the proof of this isquite similar to the triangle case,

Theorem 3.9. ∆+F (G) ≤ µF (G) with χ(F ) ≥ 3 and equality holds iff G is F−free.

For the case F = Kk(k ≥ 3) and G contains a k−clique, the equality does not hold,indeed on these conditions

Theorem 3.10. ∆+F (G) ≤ µF (G)− (

k2

)+ 2

The following conjectures [9] fit to our topic

Conjecture∆K3(G) ≤ µK3(G)

This result would be a generalization of [16] stating that the edges of a graph on n verticescan be covered by at most bn2

4c pairwise disjoint edges and triangles.

Conjecture∆F (G) ≤ µF (G)

31

if χ(F ) ≥ 3?

32

Bibliography

[1] A. Hajnal and E. Szemeredi, Proof of a conjecture of Erdos, Combinatorial theoryand its applications, Coll. Math. Soc. J. Bolyai 4, North-Holland, Amsterdam(1970),601-623

[2] B. Bollobas, On complete subgraphs of different orders, Math. Proc. CambridgePhilos Soc., 79(1976), 19-24.

[3] B. Bollobas, Extremal graph theory, Academic Press, 1978.

[4] B. Bollobas, Modern graph theory, Graduate texts in mathematics, vol 184,Springer, New York, 2002

[5] B. Sudakov and P. Keevash, Packing triangles in a Graph and its complement, J.Graph Theory 47 (2004), 203-216.

[6] E. Gyori and Zs.Tuza, decompositions of graphs into complete subgraphs of givenorder, Studia Scientiarum Mathematicarum Hungarica 22(1987), 315-320.

[7] E. Gyori, On the number of edge disjoint triangles in graphs of given size, ColloquiaMathematica Societatis Janos Bolyai, 52(1987).

[8] E. Gyori, On the number of edge disjoint cliques in graphs of given size, Combina-torica 11(1991), 231-243.

[9] J. Lehel and Zs. Tuza, Triangle-free graphs and edge covering theorems, DiscreteMathematics 39(1982), 59-65.

[10] J. Spencer, Maximal consistent families of triplets, J. Combinatorial Theory5(1968), 1-8.

[11] J. W. Moon and L.Moser, On a problem of Turan, Publ. Math. Inst. Hungar.Acad. Sci, 7(1962), 283-286.

[12] L. Lovasz and M. Simonovits, On the number of complete subgraphs of a graphII, Studies in Pure Mathematics, to the memory of P. Turan, 1983, 459-495.

33

[13] L. Lovasz The rank of connection matrices and the dimension of graph algebras,Technical report.

[14] M. Simonovits and J. Komlos, Szemeredi’s regularity lemma and its applicationsin graph theory. Combinatorics, Paul Erdos is eighty, Vol. 2 (Keszthely, 1993), 295–352, Bolyai Soc. Math. Stud., 2, Janos Bolyai Math. Soc., Budapest, 1996.

[15] N. Alon and R. Yuster, h-factors in dense graphs, J. Combin. Theory B 66(1996),269-282.

[16] P. Erdos, A. Goodman and L. Posa, The representation of graphs by set inter-sections, Canad. J. Math. 18(1966), 106-112.

[17] P. Erdos, R.J. Faudree, R.J. Gould, M.S. Jacobson, J. Lehel, Edge disjointmonochromatic triangles in 2-colored graphs. Discrete Mathematics 231(2001), 135-141.

[18] V. Rodl, On a packing and covering problem, Europ. J. Combinatorics 5(1985),69-78.

[19] V. Rodl and P. Haxell, Integer and fractional packings in dense graphs, Combina-torica 21(2001), 13-38.

34